Let $X$ be a topological space. Let $\mathcal{D}$ be a cochain algebra, with cohomology ring $H^* (\mathcal{D})$. Suppose that we have that the cohomology ring $H^* (X)$ (endowed with the cup product) is isomorphic, as a ring, to $H^* (\mathcal{D})$, equipped with its respective operation (and the homology groups of $X$ are isomorphic to the homology groups of $\mathcal{D}$).

There is an action of $H^* (X)$ on $H_* (X)$ given by the cap product: $\frown:H_* (X) \otimes H^* (X) \rightarrow H_* (X)$. There is also an action of $H^* (\mathcal{D})$ on $H_* (\mathcal{D})$ given by dualizing the algebra structure of $\mathcal{D}$ (described below). I am interested in whether this "cap product" on $\mathcal{D}$ induces the same action of $H^* (\mathcal{D})$ on $H_* (\mathcal{D})$ as the cap product does for the cohomology and homology of $X$.

Denote by $*$ the multiplication in $\mathcal{D}$. Let $\widehat{\mathcal{D}}$ denote the chain complex obtained by dualizing the cochain complex $\mathcal{D}$. We obtain an operation

$$\frown:\mathcal{D} \times \widehat{\mathcal{D}} \rightarrow \widehat{\mathcal{D}}$$

by defining $\langle u * v, w \rangle = \langle u, v \frown w \rangle$, the dual of $*$.

This is the exact way in which, if we were given the cup product on the level of cochains of $X$, we would dualize to get the cap product on the cochains/chains of $X$.

There is an induced operation on homology:

$$\frown:H_* (\mathcal{D}) \otimes H^* (\mathcal{D}) \rightarrow H_* (\mathcal{D}).$$

Is this operation the same (isomorphic to) the cap product on the isomorphic cohomology and homology groups of $X$? I think that the answer is yes, but I cannot find a proof.

EDIT: Additionally, I would at least like to know if I am correct in thinking that this is always true in the case of zero torsion.

  • $\begingroup$ Do you have non-trivial examples where this is true? I would honestly be surprised if it worked in general. $\endgroup$ Jul 2, 2020 at 13:58
  • 1
    $\begingroup$ Is this not always true if I have zero torsion? So that cup product on the level of cohomology completely determines cap? $\endgroup$
    – Matt
    Jul 4, 2020 at 20:22


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